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1. Triangle ABC has side lengths AB=9, AC=10, and BC=17. Let X be the intersection of the angle bisector of angle A with side $$\overline{BC}$$, and let Y be the foot of the perpendicular from X to side $$\overline{AC}$$. Compute the length of $$\overline{XY}$$

2. $$\overline{BC}$$ is a chord of a circle with center O and area $$48\pi$$. Point A is inside $$\triangle BCO$$ such that $$\triangle ABC$$ is equilateral and A is the circumcenter of $$\triangle BCO$$. What is the area of triangle ABC?

Thank you!!!

Apr 6, 2019
edited by Melody  Apr 7, 2019

#1
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First one....see the image below :

Let A = (0,0)

Let C = (10,0)

And we can find B, thusly

Construct a circle with radius 9 centered at the origin

The equation of this circle will be  x^2 + y^2 = 81      (1)

And construct a circle wilth radius 17 centered at (10, 0)

The equation of this circle will be  (x - 10)^2 + y^2 = 289  (2)

We can find the coordinates for point B, thusly :

Subtract  (1) from (2)  and we have that

(x-10)^2 - x^2  = 208    simplify

x^2 - 20x + 100 - x^2  = 208

-20x = 108

x = -108/20  = -5.4 =   x coordinate of B

And taking the positive value for y, we have that

(-5.4)^2 + y^2 = 81

y^2 = 81 - 29.16

y = 7.2  =  y coordinate of B

So....B =  (-5.4, 7.2)

And  AX is the bisector of angle A

So....we have the following relationship

AC/AB  = XC /XB

10/9  = XC / XB

Then BC has 19 equal parts and XC is 10 of them.....so  XC = (10/19)*17 = 170/19

So....using a similar triangle idea.....

Distance from B to the x axis / BC  =  XY / XC

7.2 / 17  = XY / (170/19)

(170/19) * 7.2 /17  = XY  =  (170/17)(7.2 / 19) = 10 * 7.2 / 19    =   72 / 19

Apr 6, 2019
#2
+106539
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Second one.....see the following image....

Let O = (0,0)

Since the area of circle O = 48pi....then the radius  = √48  units

So...the equation of the circle is x^2 + y^2 = 48

Since ABC is equilateral....then angle BAC = 60°

Then the excluded angle to  BAC = 300°

So, by symmetry, angle  AOC = 150°

And since A is the circumcenter of triangle BOC, then AO = AC = the side of equilateral triangle ABC = s

And we can find s^2 using the Law of Cosines, thusly :

OC^2 = 2s^2 - 2s^2 * ( cos 150)

48 = 2s^2 + 2s^2* (√3/2)

48 = 2s^2 + s^2√3

48 = (2 + √3)s^2

48 / [ 2 + √3 )  = s^2    = 48 [ 2 - √3]  =  96 - 48√3

So....the area of equilateral triangle ABC  =

(1/2) s^2 sin (BAC)  =

(1/2)s^2 sin(60°)  =

(√3) [ 96 - 48√3 ] / 4  =

(√3) [ 24 - 12√3]  =

24√3 - 36   units^2

Apr 7, 2019